A quadratic function has a vertex at (2,5) and passes through the point (4,9).

1 answer:

4 0

Since another way to write an equation for a parabola is a(x-h)^2-k, with h and k as the vertex points, we can plug that in to get a(x-2)^2-5. To find a, we plug 4 in and have 9 as what it equals, getting a(4-2)^2-5=9=4a-5=9 and 4a=14, where a=14/4. Therefore, the equation would be (14/4)(x-2)^2-5

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Step-by-step explanation:

Given equation: $x^2-4x+2=0$

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To find the roots, use the quadratic formula:

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$\begin{aligned}\implies x & =\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(2)}}{2(1)}\\& =\dfrac{4 \pm \sqrt{8}}{2}\\& =\dfrac{4 \pm 2\sqrt{2}}{2}\\& =2 \pm \sqrt{2}\end{aligned}$

$\textsf{Let }a=2+\sqrt{2}$

$\textsf{Let }b=2-\sqrt{2}$

Therefore:

$\begin{aligned}\implies \dfrac{a}{b^2}+\dfrac{b}{a^2} & = \dfrac{2+\sqrt{2}}{(2-\sqrt{2})^2}+\dfrac{2-\sqrt{2}}{(2+\sqrt{2})^2}\\\\& = \dfrac{2+\sqrt{2}}{6-4\sqrt{2}}+\dfrac{2-\sqrt{2}}{6+4\sqrt{2}}\\\\& = \dfrac{(2+\sqrt{2})(6+4\sqrt{2})+(2-\sqrt{2})(6-4\sqrt{2})}{(6-4\sqrt{2})(6+4\sqrt{2})}\\\\& = \dfrac{12+8\sqrt{2}+6\sqrt{2}+8+12-8\sqrt{2}-6\sqrt{2}+8}{36+24\sqrt{2}-24\sqrt{2}-32}\\\\& = \dfrac{40}{4}\\\\& = 10\end{aligned}$